We can use the Pythagorean theorem to solve for the width of the TV, since we know the height and diagonal length. Let the width be x. Then, we have:
(25)^2 + x^2 = (38)^2
625 + x^2 = 1444
x^2 = 1444 - 625
x^2 = 819
x ≈ 28.6 inches (rounded to the nearest tenth)
Therefore, the TV is approximately 28.6 inches wide.
A certain television is advertised as a 38-inch TV (the diagonal length). If the height of the TV is 25 inches, how wide is the TV? Round to the nearest tenth of an inch.
3 answers
An altitude is drawn from the vertex of an isosceles triangle, forming a right angle and two congruent triangles. As a result, the altitude cuts the base into two equal segments. The length of the altitude is 13 inches, and the length of the base is 8 inches. Find the triangle’s perimeter. Round to the nearest tenth of an inch.
Since the altitude forms two congruent triangles, each of those triangles is a right triangle with one leg of length 4 (half the length of the base) and a hypotenuse of length 13. Using the Pythagorean theorem, we can solve for the other leg of each right triangle:
(leg)^2 + 4^2 = 13^2
(leg)^2 + 16 = 169
(leg)^2 = 153
(leg) ≈ 12.4
So each leg of the isosceles triangle has length approximately 12.4 inches. The perimeter of the triangle is then:
perimeter = 2(leg) + base
perimeter = 2(12.4) + 8
perimeter ≈ 32.8 inches (rounded to the nearest tenth)
Therefore, the perimeter of the triangle is approximately 32.8 inches.
(leg)^2 + 4^2 = 13^2
(leg)^2 + 16 = 169
(leg)^2 = 153
(leg) ≈ 12.4
So each leg of the isosceles triangle has length approximately 12.4 inches. The perimeter of the triangle is then:
perimeter = 2(leg) + base
perimeter = 2(12.4) + 8
perimeter ≈ 32.8 inches (rounded to the nearest tenth)
Therefore, the perimeter of the triangle is approximately 32.8 inches.
3 answers